Prove that if $d\mid a$ and $d\mid b$ then $d\mid a\pm b$

Question

My attempt so far is that if $d\mid a$ and $d\mid b$ then $a=dx$ and $b=dy$ for some integers $x$ and $y$. Then squaring both $a$ and $b$ and taking there difference I get $a^2-b^2=d(dx^2-dy^2)$ so $(a+b)(a-b)=d(dx^2-dy^2)$

So I know that $d\mid(a+b)(a-b)$ which I'm not sure if I can use that again to help get to the end or if this was the wrong approach.

Answer 1

If $a=dx$ and $b=dy$ then $a+b= d(x+y)$ and $a-b= d(x-y).$

Answer 2

You did virtually all the work in your first sentence:

"My attempt so far is that if $d\mid a$ and $d\mid b$ then $a=dx$ and $b=dy$ for some integers $x$ and $y$."

Indeed, we then have that $a+b= dx+dy = d(x+y)$, and $a-b = dx-dy= d(x-y).$

And so, by definition, be see that $d\mid (a+b),$ and $d\mid (a-b)$.

(note: Squaring each side of each equations in this case just complicates your task.)